Two experimental runs are performed to determine the calorimetric properties of an alcohol which has a melting point of -10.0° C.In the first run,a 200-g cube of frozen alcohol,at the melting point,is added to 300 g of water at 20.0°C in a styrofoam container.When thermal equilibrium is reached,the alcohol-water solution is at a temperature of 5.0°C.In the second run,an identical cube of alcohol is added to 500 g of water at 20.0°C and the temperature at thermal equilibrium is 10.0°C.The specific heat capacity of water is 4190 J/kg ∙ K.Assume no heat is exchanged with the styrofoam container and the surroundings.What is the heat of fusion of the alcohol? A)5.5 × 10⁴ J/kg B)6.3 × 10⁴ J/kg C)7.1 × 10⁴ J/kg D)7.9 × 10⁴ J/kg E)8.7 × 10⁴ J/kg

Question

Quizplus · Accepted Answer

First, we need to determine the amount of heat that is gained or lost by the water in each experimental run using the equation:
q = mCΔT
where q is the heat gained or lost, m is the mass of the water, C is the specific heat capacity of water, and ΔT is the change in temperature of the water.

For the first run:
q = (300 g)(4190 J/kg ∙ K)(5.0°C) = 6.29 × 10^5 J
This heat comes from the melting of the alcohol, so we can use the equation:
q = mΔHfus
where ΔHfus is the heat of fusion of the alcohol and m is the mass of the alcohol that melted. Since the entire cube of alcohol melted, m = 200 g.

So, ΔHfus = q/m = (6.29 × 10^5 J)/(200 g) = 3.15 × 10^3 J/g

For the second run:
q = (500 g)(4190 J/kg ∙ K)(10.0°C) = 2.10 × 10^6 J
Again, this heat comes from the melting of the alcohol, so ΔHfus = q/m = (2.10 × 10^6 J)/(200 g) = 1.05 × 10^4 J/g

The average heat of fusion is then:
(3.15 × 10^3 J/g + 1.05 × 10^4 J/g)/2 = 6.675 × 10^3 J/g

Therefore, the best answer is B: 6.3 × 10^3 J/kg.

Two Experimental Runs Are Performed to Determine the Calorimetric Properties